我想定义一个递归函数来合并两个已排序的列表(这两个列表已排序)并返回一个新列表,其中包含两个参数列表中具有递增顺序的所有值。我知道我可以使用list.extend()和sorted()来获取它,但我不想使用它们。我只是想对递归做一些练习。
例如:
if a = [1,2,3,4], b = [5,6,7,8]
print(function(a,b))
[1,2,3,4,5,6,7,8]
这是我的代码:
def combine(a:list, b:list):
alist = []
if a == [] and b == []:
return alist
if a != [] and b == []:
return alist + a
if a == [] and b != []:
return alist + b
if a != [] and b != []:
if a[0] <= b[0]:
alist.append(a[0])
return combine(a[1:], b)
if a[0] > b[0]:
alist.append(b[0])
return combine(a, b[1:])
return alist
我总是[5,6,7,8]。如何获得[1,2,3,4,5,6,7,8]?
答案 0 :(得分:4)
您应该将其添加到alist中,而不是返回,如下所示。
def combine(a, b):
alist = []
if a == [] and b == []:
return alist
if a != [] and b == []:
return alist + a
if a == [] and b != []:
return alist + b
if a != [] and b != []:
if a[0] <= b[0]:
alist.append(a[0])
alist = alist + combine(a[1:], b)
if a[0] > b[0]:
alist.append(b[0])
alist = alist + combine(a, b[1:])
return alist
答案 1 :(得分:4)
只是一个更简单的版本:
def combine(a, b):
if a and b:
if a[0] > b[0]:
a, b = b, a
return [a[0]] + combine(a[1:], b)
return a + b
测试:
>>> combine([1,3,6,8], [2,4,5,7])
[1, 2, 3, 4, 5, 6, 7, 8]
答案 2 :(得分:1)
def combine(a,b):
if not a and not b: return []
if not a: return [b[0]] + combine(a, b[1:])
if not b: return [a[0]] + combine(a[1:], b)
if a[0] > b[0]:
return [b[0]] + combine(a, b[1:])
return [a[0]] + combine(a[1:], b)
您的测试用例:
In [2]: a = [1,2,3,4]
In [3]: b = [5,6,7,8]
In [4]: combine(a,b)
Out[4]: [1, 2, 3, 4, 5, 6, 7, 8]
另一个测试案例:
In [24]: a
Out[24]: [1, 2, 3, 8, 9]
In [25]: b
Out[25]: [1, 3, 5, 6, 7]
In [26]: combine(a,b)
Out[26]: [1, 1, 2, 3, 3, 5, 6, 7, 8, 9]
答案 3 :(得分:0)
以下是一些替代方案:
更聪明的方法是使用heapq模块中的merge函数:
from heapq import merge
list(merge(a,b))
测试:
>>> a = [1,2,3,4,7,9]
>>> b = [5,6,7,8,12]
>>> [1, 2, 3, 4, 5, 6, 7, 7, 8, 9, 12]
不使用递归:
def combine(a:list, b:list):
alist = []
i,j = 0,0
while i < len(a) and j < len(b):
if a[i] < b[j]:
alist.append(a[i])
i+=1
else:
alist.append(b[j])
j+=1
while i < len(a):
alist.append(a[i])
i+=1
while j < len(b):
alist.append(b[j])
j+=1
return alist
测试:
>>> a = [1,2,3,4,7,9]
>>> b = [5,6,7,8,12]
>>> [1, 2, 3, 4, 5, 6, 7, 7, 8, 9, 12]